When I studied physics in junior high and high school, we always took it for granted that potential energy was equal to kinetic energy. In Lagrangian terms, $T = V$at least on average. But I realized that assumption might not always be true. I have been studying the Euler-Lagrange equation and been very confused by it. But I read in Gravity by James Hartley that Newton's law can be expressed as the Euler-Lagrange equation.

My Question

If a Lagrangian satisfies the Euler-Lagrange equation, does that mean potential energy equals kinetic energy?


In other words, suppose I have a mass held above the surface of the earth with $PE = mgh$. That's the amount of energy that upon release gradually gets converted to kinetic energy. So therefore $PE = KE$ in that case. That's what I mean. I am saying I have always assumed the system I am considering is conservative. So when I looked at the Euler Lagrange equation, it looked trivial because had (without knowing) assumed every system satisfied it. So I am trying to see if I have the story straight this time.

  • 1
    $\begingroup$ You ask your question in a wrong way: when a body moves in a conservative field, the energy transforms from potential to kinetic and vice-versa. But, at a given time, the two energies are not necessarily equal. So, first, I suggest you to correct your question, and then it can be more easily examined. $\endgroup$
    – Sofia
    Jan 22, 2015 at 22:11
  • $\begingroup$ The phrase "gauge transformation" may be apropos here. In the usual gauge choice two-body orbital mechanics has the property you suggest, but ... $\endgroup$ Jan 22, 2015 at 22:21
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    $\begingroup$ But $T \neq V$ in general. However $dT/dt = -dV/dt$ in general. $\endgroup$
    – Jold
    Jan 22, 2015 at 23:05
  • $\begingroup$ @jld maybe that's part of why i am confused. Why can't $T=V$? I get that it might always not but are you saying it cannot be equal ever? $\endgroup$ Jan 22, 2015 at 23:09
  • $\begingroup$ @StanShunpike $T=V$ under special circumstances. What's important is that $T+V=constant$ over time. There may be some points on a particle's path where $T=V$, but it certainly won't be true everywhere. $\endgroup$
    – Jold
    Jan 22, 2015 at 23:45

2 Answers 2


Potential energy is always measured relative to something, just as kinetic energy is measured in a particular frame of reference. That being the case, I could always make $T=V$ by choosing the reference in a certain way - but it will not generally be true.

For example, I can set $T=0$ at the top of a tower, or at ground level. Both are valid - it's really up to me. Or I can say $T=0$ at infinite distance from the earth.

And when I observe a falling object, that object will appear to be decelerating relative to a frame of reference that is moving towards the earth faster than the object.

Instead, you can say that

$$T + V = \mathrm{constant}$$

Because as you lose potential energy, you gain kinetic energy (assuming those are the only two forms that energy can take in your system). If you add friction, it's of course possible that the sum $T+V$ will decrease as the system warms up (unless you define thermal energy as a form of either potential or kinetic energy, which you could...)

But no, $T=V$ is not in general true.

  1. No, the (time-averaged) kinetic and potential energies, $\langle T \rangle$ and $\langle V \rangle$, are not the same in general.

  2. E.g. the virial theorem in Newtonian mechanics shows that $$ 2\langle T \rangle ~=~p\langle V \rangle ,$$ if the potential $$V(r) \propto r^p $$ is a power law.

  3. In particular, for a satellite in Newtonian gravity, $\langle V \rangle$ is minus two times $\langle T \rangle$.


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